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Which is the weaker sequence? (part 2)


Which is the weaker sequence? (see post)  

28 members have voted

  1. 1. Which is the weaker sequence? (see post)

    • 1d-(x)-xx-(2s)-3c
      16
    • 1d-(x)-xx-(2s)-p-(p)-x-(p)-3c
      12


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You were playing that the redouble creates a forcing auction through 2

Nice question. Instinctively one would feel that if we have the balance of strength then the higher the opps bid, the less inclined we should be to allow them to buy it undoubled.

umm? You seem to be assuming the opponents bidding is random or something. The higher they bid the more they will have in distribution/fit/overall values. Why would you want to be in a force over 2S just because you have slightly more than half the HCP when the opps know this and have bid 2S anyways. Certainly if they bid higher than that being in a force is ridiculous. You should only be forced as high as your general values would dictate, here you have 12+ opp 10+ so being forced to the 3 level is a bit silly.

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Also this should be obvious but if you are not in a force then doubles are not penalty as that would be unplayable. So the second auction is just logically:

 

PASS= minimum since you are willing to defend 2S undoubled and it's a non forcing pass.

 

DOUBLE= takeout.

 

3C= ok I took out your t/o X.

 

So you first made a non forcing pass showing no extras then took out partners t/o X to clubs.

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You were playing that the redouble creates a forcing auction through 2

Nice question. Instinctively one would feel that if we have the balance of strength then the higher the opps bid, the less inclined we should be to allow them to buy it undoubled.

umm? You seem to be assuming the opponents bidding is random or something. The higher they bid the more they will have in distribution/fit/overall values. Why would you want to be in a force over 2S just because you have slightly more than half the HCP when the opps know this and have bid 2S anyways.

I suppose because one of two mutually exclusive scenarious apply: Either they are (1) in a fit xor (2) they are not. If they are in a misfit and trying for 8 tricks to our 5 with less than half the values, they should be going down, at least sufficiently frequently that catering for the possibility of their making is likely to be a waste of resources. Conversely, if both sides have a fit, and we have the balance of the strength, it still does not seem right to defend undoubled. If they are making then we should be saving. Either way it doesn't sound right to pass it out.

 

Obviously exceptional counter-examples can be constructed, but the question again arises whether it is worthwhile devoting resources to cater for the possibility.

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